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Prove if $f(0) = 0$ then $\displaystyle\lim_{x \to 0^+}x\int_x^1 \frac{f(t)}{t^2}dt = 0$ for regulated function $f$

A regulated function is a function $f$ on $[a,b]$ such that $\exists$ a sequence $(\varphi_n)_{n \in \mathbb{N}}$ of step functions such that $\displaystyle \lim_{n \to \infty} \sup_{x \in [a,b]} \lvert f(x) - \varphi_n(x) \rvert = 0$ and $\forall x \in (a,b)$ the left and right limits exist, also left limit of $b$ & right limit of $a$. Also given the assumption that $f$ is continuous at $0$ which is said to be redundant.

So we know $f$ is continuous on $[0,1)$. For $\varepsilon > 0, \exists \delta > 0, \lvert x \rvert < \delta \implies \lvert f(x) - f(0) \rvert = \lvert f(x) \rvert < \varepsilon$

Intuitively, I know the $x$ outside the integral goes to $0$ so as long as the integral itself converges then we can get the desired $0$ as the limit.

so get $\displaystyle\lim_{x \to 0^+}x \cdot \lim_{x \to 0^+}\int_x^1 \frac{f(t)}{t^2}dt = 0 \cdot \lim_{x \to 0^+}\int_x^1 \frac{f(t)}{t^2}dt$ so we need $\displaystyle\lim_{x \to 0^+}\int_x^1 \frac{f(t)}{t^2}dt < \infty$

Now I'm stuck because if $x = 0$ then $\displaystyle \frac{f(0)}{0} = \frac{0}{0}$

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    $\begingroup$ You may not want to "break" the limit in two -- you do need the factor $x$ for things to go well. To see why, take the function $f\colon x\mapsto x$. It satisfies your assumptions, but you cannot really say that $\int_x^1 \frac{f(t)}{t^2}dt = -\ln x$ will converge nicely to anything finite. $\endgroup$
    – Clement C.
    Commented Jan 17, 2017 at 3:14

1 Answer 1

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Let $\epsilon>0$ be given and choose $\delta >0$ so small that $|f(x)-f(0)|=|f(x)|<\epsilon$ for all $x\in (0,\delta)$.

Then, we can write

$$x\int_x^1\frac{f(t)}{t^2}\,dt=x\int_x^\delta \frac{f(t)}{t^2}\,dt+x\int_\delta^1 \frac{f(t)}{t^2}\,dt \tag 1$$

With this $\delta$ fixed, the second integral on the right-hand side of $(1)$ goes to zero as $x\to 0$. For the first integral we have

$$\left|x\int_x^\delta \frac{f(t)}{t^2}\,dt\right|\le x\int_x^\delta \frac{|f(t)|}{t^2}\,dt<\epsilon \left(1-\frac x\delta\right)$$

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  • $\begingroup$ The second integral on the right side of (1) goes to $0$ because $x \to 0$ and $\frac{f(t)}{t^2} \not \to \frac{0}{0}$, correct? $\endgroup$ Commented Jan 17, 2017 at 3:41
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    $\begingroup$ @oliverjones Yes, $x$ goes to $0$ and the integral is fixed (does not change as $x$ varies: it is a number, which stays what it is). $\endgroup$
    – Clement C.
    Commented Jan 17, 2017 at 4:01
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    $\begingroup$ Another approach is to use "good ole" L'Hospital's Rule. We can write $$\lim_{x\to 0}x\int_x^1 \frac{f(t)}{t^2}\,dt=\lim_{x\to 0} \frac{\int_x^1 \frac{f(t)}{t^2}\,dt}{1/x}=\lim_{x\to 0} \frac{-\frac{f(x)}{x^2}}{-1/x^2}=\lim_{x\to 0}f(x)=f(0)=0$$ $\endgroup$
    – Mark Viola
    Commented Jan 17, 2017 at 14:36

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